3.3.30 \(\int \frac {x \sqrt {a+c x^2}}{d+e x} \, dx\)

Optimal. Leaf size=127 \[ \frac {\left (a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c} e^3}+\frac {d \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^3}-\frac {\sqrt {a+c x^2} (2 d-e x)}{2 e^2} \]

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Rubi [A]  time = 0.11, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {815, 844, 217, 206, 725} \begin {gather*} \frac {\left (a e^2+2 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c} e^3}+\frac {d \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{e^3}-\frac {\sqrt {a+c x^2} (2 d-e x)}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*Sqrt[a + c*x^2])/(d + e*x),x]

[Out]

-((2*d - e*x)*Sqrt[a + c*x^2])/(2*e^2) + ((2*c*d^2 + a*e^2)*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]])/(2*Sqrt[c]*e
^3) + (d*Sqrt[c*d^2 + a*e^2]*ArcTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/e^3

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {x \sqrt {a+c x^2}}{d+e x} \, dx &=-\frac {(2 d-e x) \sqrt {a+c x^2}}{2 e^2}+\frac {\int \frac {-a c d e+c \left (2 c d^2+a e^2\right ) x}{(d+e x) \sqrt {a+c x^2}} \, dx}{2 c e^2}\\ &=-\frac {(2 d-e x) \sqrt {a+c x^2}}{2 e^2}-\frac {\left (d \left (c d^2+a e^2\right )\right ) \int \frac {1}{(d+e x) \sqrt {a+c x^2}} \, dx}{e^3}+\frac {\left (2 c d^2+a e^2\right ) \int \frac {1}{\sqrt {a+c x^2}} \, dx}{2 e^3}\\ &=-\frac {(2 d-e x) \sqrt {a+c x^2}}{2 e^2}+\frac {\left (d \left (c d^2+a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{c d^2+a e^2-x^2} \, dx,x,\frac {a e-c d x}{\sqrt {a+c x^2}}\right )}{e^3}+\frac {\left (2 c d^2+a e^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x}{\sqrt {a+c x^2}}\right )}{2 e^3}\\ &=-\frac {(2 d-e x) \sqrt {a+c x^2}}{2 e^2}+\frac {\left (2 c d^2+a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )}{2 \sqrt {c} e^3}+\frac {d \sqrt {c d^2+a e^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {c d^2+a e^2} \sqrt {a+c x^2}}\right )}{e^3}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 175, normalized size = 1.38 \begin {gather*} \frac {\frac {a^{3/2} e^2 \sqrt {\frac {c x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {c} \sqrt {a+c x^2}}+2 d \sqrt {a e^2+c d^2} \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )+2 \sqrt {c} d^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a+c x^2}}\right )-2 d e \sqrt {a+c x^2}+e^2 x \sqrt {a+c x^2}}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x*Sqrt[a + c*x^2])/(d + e*x),x]

[Out]

(-2*d*e*Sqrt[a + c*x^2] + e^2*x*Sqrt[a + c*x^2] + (a^(3/2)*e^2*Sqrt[1 + (c*x^2)/a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]
])/(Sqrt[c]*Sqrt[a + c*x^2]) + 2*Sqrt[c]*d^2*ArcTanh[(Sqrt[c]*x)/Sqrt[a + c*x^2]] + 2*d*Sqrt[c*d^2 + a*e^2]*Ar
cTanh[(a*e - c*d*x)/(Sqrt[c*d^2 + a*e^2]*Sqrt[a + c*x^2])])/(2*e^3)

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IntegrateAlgebraic [A]  time = 0.47, size = 177, normalized size = 1.39 \begin {gather*} \frac {\left (-a e^2-2 c d^2\right ) \log \left (\sqrt {a+c x^2}-\sqrt {c} x\right )}{2 \sqrt {c} e^3}-\frac {2 d \sqrt {-a e^2-c d^2} \tan ^{-1}\left (-\frac {e \sqrt {a+c x^2}}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} d}{\sqrt {-a e^2-c d^2}}\right )}{e^3}+\frac {\sqrt {a+c x^2} (e x-2 d)}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x*Sqrt[a + c*x^2])/(d + e*x),x]

[Out]

((-2*d + e*x)*Sqrt[a + c*x^2])/(2*e^2) - (2*d*Sqrt[-(c*d^2) - a*e^2]*ArcTan[(Sqrt[c]*d)/Sqrt[-(c*d^2) - a*e^2]
+ (Sqrt[c]*e*x)/Sqrt[-(c*d^2) - a*e^2] - (e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]])/e^3 + ((-2*c*d^2 - a*e^
2)*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2]])/(2*Sqrt[c]*e^3)

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fricas [A]  time = 0.68, size = 684, normalized size = 5.39 \begin {gather*} \left [\frac {2 \, \sqrt {c d^{2} + a e^{2}} c d \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + {\left (2 \, c d^{2} + a e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (c e^{2} x - 2 \, c d e\right )} \sqrt {c x^{2} + a}}{4 \, c e^{3}}, \frac {4 \, \sqrt {-c d^{2} - a e^{2}} c d \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) + {\left (2 \, c d^{2} + a e^{2}\right )} \sqrt {c} \log \left (-2 \, c x^{2} - 2 \, \sqrt {c x^{2} + a} \sqrt {c} x - a\right ) + 2 \, {\left (c e^{2} x - 2 \, c d e\right )} \sqrt {c x^{2} + a}}{4 \, c e^{3}}, \frac {\sqrt {c d^{2} + a e^{2}} c d \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} + 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - {\left (2 \, c d^{2} + a e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (c e^{2} x - 2 \, c d e\right )} \sqrt {c x^{2} + a}}{2 \, c e^{3}}, \frac {2 \, \sqrt {-c d^{2} - a e^{2}} c d \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) - {\left (2 \, c d^{2} + a e^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x}{\sqrt {c x^{2} + a}}\right ) + {\left (c e^{2} x - 2 \, c d e\right )} \sqrt {c x^{2} + a}}{2 \, c e^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^2+a)^(1/2)/(e*x+d),x, algorithm="fricas")

[Out]

[1/4*(2*sqrt(c*d^2 + a*e^2)*c*d*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2)*x^2 + 2*sqrt(c*
d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) + (2*c*d^2 + a*e^2)*sqrt(c)*log(-2*c*x^
2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(c*e^2*x - 2*c*d*e)*sqrt(c*x^2 + a))/(c*e^3), 1/4*(4*sqrt(-c*d^2 - a*
e^2)*c*d*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^
2)) + (2*c*d^2 + a*e^2)*sqrt(c)*log(-2*c*x^2 - 2*sqrt(c*x^2 + a)*sqrt(c)*x - a) + 2*(c*e^2*x - 2*c*d*e)*sqrt(c
*x^2 + a))/(c*e^3), 1/2*(sqrt(c*d^2 + a*e^2)*c*d*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e^2 - (2*c^2*d^2 + a*c*e^2
)*x^2 + 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2)) - (2*c*d^2 + a*e^2)*sq
rt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (c*e^2*x - 2*c*d*e)*sqrt(c*x^2 + a))/(c*e^3), 1/2*(2*sqrt(-c*d^2 -
a*e^2)*c*d*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)
*x^2)) - (2*c*d^2 + a*e^2)*sqrt(-c)*arctan(sqrt(-c)*x/sqrt(c*x^2 + a)) + (c*e^2*x - 2*c*d*e)*sqrt(c*x^2 + a))/
(c*e^3)]

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giac [A]  time = 0.20, size = 135, normalized size = 1.06 \begin {gather*} -\frac {2 \, {\left (c d^{3} + a d e^{2}\right )} \arctan \left (-\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right ) e^{\left (-3\right )}}{\sqrt {-c d^{2} - a e^{2}}} - \frac {{\left (2 \, c^{\frac {3}{2}} d^{2} + a \sqrt {c} e^{2}\right )} e^{\left (-3\right )} \log \left ({\left | -\sqrt {c} x + \sqrt {c x^{2} + a} \right |}\right )}{2 \, c} + \frac {1}{2} \, \sqrt {c x^{2} + a} {\left (x e^{\left (-1\right )} - 2 \, d e^{\left (-2\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^2+a)^(1/2)/(e*x+d),x, algorithm="giac")

[Out]

-2*(c*d^3 + a*d*e^2)*arctan(-((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^(-3)/sqrt(-
c*d^2 - a*e^2) - 1/2*(2*c^(3/2)*d^2 + a*sqrt(c)*e^2)*e^(-3)*log(abs(-sqrt(c)*x + sqrt(c*x^2 + a)))/c + 1/2*sqr
t(c*x^2 + a)*(x*e^(-1) - 2*d*e^(-2))

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maple [B]  time = 0.01, size = 423, normalized size = 3.33 \begin {gather*} \frac {a d \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{2}}+\frac {c \,d^{3} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, e^{4}}+\frac {a \ln \left (\sqrt {c}\, x +\sqrt {c \,x^{2}+a}\right )}{2 \sqrt {c}\, e}+\frac {\sqrt {c}\, d^{2} \ln \left (\frac {-\frac {c d}{e}+\left (x +\frac {d}{e}\right ) c}{\sqrt {c}}+\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\right )}{e^{3}}+\frac {\sqrt {c \,x^{2}+a}\, x}{2 e}-\frac {\sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, d}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(c*x^2+a)^(1/2)/(e*x+d),x)

[Out]

1/2/e*x*(c*x^2+a)^(1/2)+1/2/e*a/c^(1/2)*ln(c^(1/2)*x+(c*x^2+a)^(1/2))-d/e^2*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e
^2+c*d^2)/e^2)^(1/2)+d^2/e^3*c^(1/2)*ln((-c*d/e+(x+d/e)*c)/c^(1/2)+(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)
/e^2)^(1/2))+d/e^2/((a*e^2+c*d^2)/e^2)^(1/2)*ln((-2*(x+d/e)*c*d/e+2*(a*e^2+c*d^2)/e^2+2*((a*e^2+c*d^2)/e^2)^(1
/2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^2)^(1/2))/(x+d/e))*a+d^3/e^4/((a*e^2+c*d^2)/e^2)^(1/2)*ln((-
2*(x+d/e)*c*d/e+2*(a*e^2+c*d^2)/e^2+2*((a*e^2+c*d^2)/e^2)^(1/2)*(-2*(x+d/e)*c*d/e+(x+d/e)^2*c+(a*e^2+c*d^2)/e^
2)^(1/2))/(x+d/e))*c

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maxima [A]  time = 0.51, size = 122, normalized size = 0.96 \begin {gather*} \frac {\sqrt {c x^{2} + a} x}{2 \, e} + \frac {\sqrt {c} d^{2} \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{e^{3}} + \frac {a \operatorname {arsinh}\left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {c} e} - \frac {\sqrt {a + \frac {c d^{2}}{e^{2}}} d \operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{e^{2}} - \frac {\sqrt {c x^{2} + a} d}{e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x^2+a)^(1/2)/(e*x+d),x, algorithm="maxima")

[Out]

1/2*sqrt(c*x^2 + a)*x/e + sqrt(c)*d^2*arcsinh(c*x/sqrt(a*c))/e^3 + 1/2*a*arcsinh(c*x/sqrt(a*c))/(sqrt(c)*e) -
sqrt(a + c*d^2/e^2)*d*arcsinh(c*d*x/(sqrt(a*c)*abs(e*x + d)) - a*e/(sqrt(a*c)*abs(e*x + d)))/e^2 - sqrt(c*x^2
+ a)*d/e^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,\sqrt {c\,x^2+a}}{d+e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + c*x^2)^(1/2))/(d + e*x),x)

[Out]

int((x*(a + c*x^2)^(1/2))/(d + e*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x \sqrt {a + c x^{2}}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(c*x**2+a)**(1/2)/(e*x+d),x)

[Out]

Integral(x*sqrt(a + c*x**2)/(d + e*x), x)

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